মনে করি,
\(x\)-এর অতিক্ষুদ্র বৃদ্ধি \(\delta x\)-এর জন্য, \(y\) ও \(z\) -এ বৃদ্ধি যথাক্রমে \(\delta y\) এবং \(\delta z\);
যেহেতু এই বৃদ্ধিগুলি সসীম,
অতএব,
\(\frac{\delta y}{\delta x}=\frac{\delta y}{\delta z}.\frac{\delta z}{\delta x}\)
\[\therefore \lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}=\lim_{\delta x\rightarrow 0}\left(\frac{\delta y}{\delta z}.\frac{\delta z}{\delta x}\right)\]
\[=\lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta z}.\lim_{\delta x\rightarrow 0}\frac{\delta z}{\delta x}\]
যেহেতু \(z, x\)-এর একটি ফাংশন,
অতএব,
\(\delta x\rightarrow 0\) হলে \(\delta z\rightarrow 0\) হবে।
\[\therefore \lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}=\lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta z}.\lim_{\delta x\rightarrow 0}\frac{\delta z}{\delta x}\]
\[\Rightarrow \frac{dy}{dx}=\frac{dy}{dz}.\frac{dz}{dx}\] ➜ Note\[\because \lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}=\frac{dy}{dx}\]
(proved)

মনে করি,
\(x\)-এর অতিক্ষুদ্র বৃদ্ধি \(\delta x\)-এর জন্য, \(y\)-এর বৃদ্ধি \(\delta y\);
যেহেতু এই বৃদ্ধিগুলি সসীম,
অতএব,
\(\frac{\delta y}{\delta x}=\frac{1}{\frac{\delta x}{\delta y}}\)
\[\therefore \lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}= \lim_{\delta x\rightarrow 0}\frac{1}{\frac{\delta x}{\delta y}}\]
\[\Rightarrow \lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}= \frac{1}{\lim_{\delta x\rightarrow 0}\frac{\delta x}{\delta y}}\]
\[\Rightarrow \frac{dy}{dx}= \frac{1}{\frac{dx}{dy}}\] ➜ Note\[\because \lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}=\frac{dy}{dx}\]
(proved)

মনে করি,
\(\sin^{-1} x=y\)
\(\Rightarrow x=\sin y\)
\(\therefore \frac{d}{dy}(x)=\frac{d}{dy}(\sin y)\) ➜ Note\(y\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\cos y\)
\(=\sqrt{1-\sin^2 y}\) ➜ Note\(\because \cos A=\sqrt{1-\sin^2 A}\)
\(=\sqrt{1-x^2}\) ➜ Note\(\because x=\sin y\)
\(\therefore \frac{dx}{dy}=\sqrt{1-x^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{1}{\sqrt{1-x^2}}\)
\(\therefore \frac{d}{dx}(\sin^{-1} x)=\frac{1}{\sqrt{1-x^2}}\)
(proved)

মনে করি,
\(\cos^{-1} x=y\)
\(\Rightarrow x=\cos y\)
\(\therefore \frac{d}{dy}(x)=\frac{d}{dy}(\cos y)\) ➜ Note\(y\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=-\sin y\)
\(=-\sqrt{1-\cos^2 y}\) ➜ Note\(\because \sin A=\sqrt{1-\cos^2 A}\)
\(=-\sqrt{1-x^2}\) ➜ Note\(\because x=\cos y\)
\(\therefore \frac{dx}{dy}=-\sqrt{1-x^2}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \frac{d}{dx}(y)=-\frac{1}{\sqrt{1-x^2}}\)
\(\therefore \frac{d}{dx}(\cos^{-1} x)=-\frac{1}{\sqrt{1-x^2}}\)
(proved)

মনে করি,
\(\tan^{-1} x=y\)
\(\Rightarrow x=\tan y\)
\(\therefore \frac{d}{dy}(x)=\frac{d}{dy}(\tan y)\) ➜ Note\(y\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\sec^2 y\)
\(=1+\tan^2 y\) ➜ Note\(\because \sec^2 A=1+\tan^2 A\)
\(=1+x^2\) ➜ Note\(\because x=\tan y\)
\(\therefore \frac{dx}{dy}=1+x^2\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{1+x^2}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{1}{1+x^2}\)
\(\therefore \frac{d}{dx}(\tan^{-1} x)=\frac{1}{1+x^2}\)
(proved)

মনে করি,
\(cosec^{-1} \ x=y\)
\(\Rightarrow x=cosec \ y\)
\(\therefore \frac{d}{dy}(x)=\frac{d}{dy}(cosec \ y)\) ➜ Note\(y\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=-cosec \ y\cot y\)
\(=-cosec \ y\sqrt{cosec^2 \ y-1}\) ➜ Note\(\because \cot A=\sqrt{cosec^2 \ A-1}\)
\(=-x\sqrt{x^2-1}\) ➜ Note\(\because x=cosec \ y\)
\(\therefore \frac{dx}{dy}=-x\sqrt{x^2-1}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{1}{x\sqrt{x^2-1}}\)
\(\Rightarrow \frac{d}{dx}(y)=-\frac{1}{x\sqrt{x^2-1}}\)
\(\therefore \frac{d}{dx}(cosec^{-1} \ x)=-\frac{1}{x\sqrt{x^2-1}}\)
(proved)

মনে করি,
\(\sec^{-1} x=y\)
\(\Rightarrow x=\sec y\)
\(\therefore \frac{d}{dy}(x)=\frac{d}{dy}(\sec y)\) ➜ Note\(y\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\sec y\tan y\)
\(=\sec y\sqrt{\sec^2 y-1}\) ➜ Note\(\because \tan A=\sqrt{\sec^2 A-1}\)
\(=x\sqrt{x^2-1}\) ➜ Note\(\because x=\sec y\)
\(\therefore \frac{dx}{dy}=x\sqrt{x^2-1}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{x\sqrt{x^2-1}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{1}{x\sqrt{x^2-1}}\)
\(\therefore \frac{d}{dx}(\sec^{-1} x)=\frac{1}{x\sqrt{x^2-1}}\)
(proved)

মনে করি,
\(\cot^{-1} x=y\)
\(\Rightarrow x=\cot y\)
\(\therefore \frac{d}{dy}(x)=\frac{d}{dy}(\cot y)\) ➜ Note\(y\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=-cosec^2 \ y\)
\(=-(1+\cot^2 y)\) ➜ Note\(\because cosec^2 \ A=1+\cot^2 A\)
\(=-(1+x^2)\) ➜ Note\(\because x=\cot y\)
\(\therefore \frac{dx}{dy}=-(1+x^2)\)
\(\Rightarrow \frac{dy}{dx}=-\frac{1}{1+x^2}\)
\(\Rightarrow \frac{d}{dx}(y)=-\frac{1}{1+x^2}\)
\(\therefore \frac{d}{dx}(\cot^{-1} x)=-\frac{1}{1+x^2}\)
(proved)